Question

If 3,-2 are the zeroes of the polymonial p(x) then find p(x)​

Answer 1

Answer࿐

Given,

3x+2y = -4 ----------(1)

2x+5y = 1 -----------(2)

¶ Find Point of Intersection of these 2 lines

Do 2×(1) - 3×(2)

6x+4y = -8 ----------(3)

6x+15y = 3 ---------(4)

-----------------------

-11y = -11

=> y = 1

Substitute in (1)

3x+2(1) = -4

=> 3x = -4-2

=> 3x = -6

=> x = -2

•°• Point of Intersection of the lines 3x+2y+4= 0 & 2x+5y-1= 0 is (-2,1)

¶ By using the Slope-Intercept form, find the form of equation of line passing through the point (-2,1)

y = mx + c

substitute x = -2 & y = 1

=> 1 = -2m + c

=> c = 1 + 2m

•°• The Required Equations of straight line is of form :

y = mx + 1 + 2m -----------(5)

¶ The perpendicular distance (or simply distance) 'd' of a point P(x1,y1) from Ax+By+C = 0 is given by

Given,

(x1,y1) = (-2,1) & d = 2

=> (4m+2)² = 2(m² + 1)

=> 16m² + 16m + 4 = 2m² + 2

=> 14m² + 16m + 2 = 0

=> 7m² + 8m + 1 = 0

Factorise the equation

=> 7m² + 7m + m + 1 = 0

=> 7m(m+1) + 1(m+1) = 0

=> (m+1)(7m+1) = 0

=> m = -1 and m = -1/7

Now Substitute m = -1 in (5)

=> y = (-1)x+1+2(-1)

=> y = -x + 1 - 2

=> y = -1 - x

(or)

=> -x - y -1 = 0

=> x + y + 1 = 0 ------------(6)

Substitute m = -1/7 in (5)

=> y = (-1/7)x + 1 + 2(-1/7)

=> y = -x/7 + (7-2)/7

=> y = (-x+5)/7

=> 7y = -x+5

or

=> -x - 7y + 5 = 0

=> x + 7y - 5 = 0 ----------(7)

•°• The Required equation of straight lines are :

x + y + 1 = 0 & x + 7y - 5 = 0

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Answer 2

${\large{\frak{\pmb{\underline{Given \; That}}}}}$

3,-2 are the zeros of the polymonial p(x).

${\large{\frak{\pmb{\underline{To \; find}}}}}$

Value of p(x)

${\large{\frak{\pmb{\underline{Solution}}}}}$

Value of p(x) = x² -x -6 = 0

${\large{\frak{\pmb{\underline{Using \; concept}}}}}$

★ Formula to find sum of the zeros of the quadratic equation

★ Formula to find product of the zeros of the quadratic equation

★ Formula of quadratic equation.

${\large{\frak{\pmb{\underline{Using \; formula}}}}}$

★ Formula to find sum of the zeros of the quadratic equation = α+β = -b/a

★ Formula to find product of the zeros of the quadratic equation = αβ = c/a

★ Formula of quadratic equation = x²-(α+β)x + αβ

${\large{\frak{\pmb{\underline{Full \; Solution}}}}}$

~ As we already know that the sum of the zeros of the quadratic equation is given by α+β = -b/a.

↝ α+β = -b/a

↝ 3+(-2)

↝ 3-2

↝ 1

• Therefore, value of α+β is 1.

~ As we already know that the product of the zeros of the quadratic equation is given by αβ = c/a

↝ αβ = c/a

↝ 3(-2)

↝ -6

• Therefore, value of αβ is -6.

~ As we already know that formula of quadratic equation is given by x²-(α+β)x + αβ

↝ x²-(α+β)x + αβ = 0

↝ x² (-1)x + (-6) = 0

↝ x² -x -6 = 0

• Therefore, x² -x -6 = 0 is the value of p(x).